The value of 'c' in Rolle's Theorem for the function f(x) =  on [π, 3π]:

Concept:

Rolle's theorem:

Rolle's theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that,

f(a) = f(b), then, for some c ∈ [a, b]

f′(c) = 0 

Calculation:

The given function is f(x) =  on [π, 3π].

f(π) =  = 0 and f(3π) =  = 0.

Since, f(π) = f(3π), there must exist a c ∈ [π, 3π] such that f'(c) = 0.

f'(x) = 

⇒ f'(c) =  = 0

⇒  = 0

⇒  = nπ

⇒ c = 2nπ, where n is an integer.

We want c ∈ [π, 3π], therefore c = 2π.